CODEWITHSUNDEEP
javascript
Konstantin
Konstantin

Reputation: 41

Array destructuring - unexpected behavior

const [a, b] = [1, 2]
// expected to be equal to 
// const a = 1
// const b = 2
const [c, d] = [3, 4]
// expected to be equal to 
// const c = 3
// const d = 4
[a, b] = [b, a] 
// expected Assignment to constant variable error
console.log(a, b, c, d)
// => 1, 2, 2, 1

Can someone explain what is wrong with this code? What happened to variables c and d?

Upvotes: 3

Views: 196

Answers (3)

Paul
Paul

Reputation: 141829

You're not using semi-colons and ASI doesn't always work the way you would like.

Your code is equivalent to this:

const [a, b] = [1, 2];

const [c, d] = [3, 4][a, b] = [b, a];

console.log(a, b, c, d);

[3, 4][a, b] in this case evaluates as [3,4][2] since the second comma is just the comma operator and b === 2. That makes the assignment equivalent to [3,4][2] = [b,a] which results in a temporary array [3,4,[2,1]] being created and the result of the assignment expression [2,1] is then assigned to [c,d].

Try adding semi-colons and you will get the expected results:

const [a, b] = [1, 2];
// expected to be equal to 
// const a = 1
// const b = 2
const [c, d] = [3, 4];
// expected to be equal to 
// const c = 3
// const d = 4
[a, b] = [b, a];
// expected Assignment to constant variable error
console.log(a, b, c, d);

Upvotes: 4

Tilepaper
Tilepaper

Reputation: 88

Instead of:

const [a, b] = [1, 2];

Try:

const a = 1, b = 2;
const c = 3, d = 4;

To swap the values, try something like:

var temp;
temp = a; a = b; b = temp;

Upvotes: -3

rorschach
rorschach

Reputation: 2947

It's basically a problem with not terminating commands properly with a semi-colon (;). If you add them to the end of each command you will get the proper result.. and also run into the problem where you cannot reassign a and b as they're declared as const.

Upvotes: 1

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